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Gekata [30.6K]
3 years ago
5

What decimals are greater than 5

Mathematics
1 answer:
malfutka [58]3 years ago
8 0
For example, 

7.2, 4.3, 13.9
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50 Points What is the value of r? 6.4r−5.9r=−3.25 Question 1 options: A -6.5 B 6.5 C -3
Oliga [24]

Answer:

Option A. -6.5

Step-by-step explanation:

we have

6.4r-5.9r=-3.25

Solve for r

That means ---> isolate the variable r

Combine like terns left side of the equation

0.5r=-3.25

Divide by 0.5 both sides

0.5r/0.5=-3.25/0.5

r=-6.5

4 0
3 years ago
one side of rectangle is 3 inches shorter than the other side, the perimeter is 54 inches. what are the dimensions of the rectan
dimaraw [331]

Answer:

2x + 2(x - 3) = 54

Please mark as brainliest


5 0
3 years ago
Read 2 more answers
The bad debt ratio for a financial institution is defined to be the dollar values of loans defaulted divided by the total dollar
Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, H_0 : \mu \leq  3.5%

    ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, H_0 : \mu \leq  3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

                T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where,  \bar X = sample mean debt ratio of Ohio banks = 6%

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 1.83%

             n = sample of banks = 7

So, test statistics = \frac{6-3.5}{\frac{1.83}{\sqrt{7} } }  ~ t_6

                             = 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

7 0
3 years ago
When d the product of 6 and 6 is divided by the sum of 6 and 6 what is the quotient? Explain.
shepuryov [24]
Hope it helps! If it is, Brainliest please!

7 0
3 years ago
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The statement “cot theta= 12/5, sec theta, and the terminal point determined by theta is in quadrant 2.”
PIT_PIT [208]

Answer: cannot be true because cot∅ is less than zero in quadrant 2.

Step-by-step explanation:

Apex trigonometry answer

7 0
3 years ago
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